Maze-hdu4035（DP求概率）

链接：http://acm.hdu.edu.cn/showproblem.php?pid=4035

题意：
有n个房间，由n-1条隧道连通起来，实际上就形成了一棵树，
从结点1出发，开始走，在每个结点i都有3种可能：
1.被杀死，回到结点1处（概率为ki）
2.找到出口，走出迷宫 （概率为ei）
3.和该点相连有m条边，随机走一条
求：走出迷宫所要走的边数的期望值。

设 E[i]表示在结点i处，要走出迷宫所要走的边数的期望。E[1]即为所求。

叶子结点：
E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);
= ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei);

非叶子结点：（m为与结点相连的边数）
E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) );
= ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei);

设对每个结点：E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;

对于非叶子结点i，设j为i的孩子结点，则
∑(E[child[i]]) = ∑E[j]
= ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
= ∑(Aj*E[1] + Bj*E[i] + Cj)
带入上面的式子得
(1 - (1-ki-ei)/m*∑Bj)*E[i] = (ki+(1-ki-ei)/m*∑Aj)*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑Cj;
由此可得
Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj);
Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj);
Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj);

对于叶子结点
Ai = ki;
Bi = 1 - ki - ei;
Ci = 1 - ki - ei;

从叶子结点开始，直到算出 A1,B1,C1;

E[1] = A1*E[1] + B1*0 + C1;
所以
E[1] = C1 / (1 - A1);
若 A1趋近于1则无解...

转载自博客：https://blog.csdn.net/morgan_xww/article/details/6776947/

代码：

#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
 
using namespace std;
 
const int MAXN = 10000 + 5;
 
double e[MAXN], k[MAXN];
double A[MAXN], B[MAXN], C[MAXN];
 
vector<int> v[MAXN];
 
bool search(int i, int fa)
{
    if ( v[i].size() == 1 && fa != -1 )
    {
        A[i] = k[i];
        B[i] = 1 - k[i] - e[i];
        C[i] = 1 - k[i] - e[i];
        return true;
    }
 
    A[i] = k[i];
    B[i] = (1 - k[i] - e[i]) / v[i].size();
    C[i] = 1 - k[i] - e[i];
    double tmp = 0;
    
    for (int j = 0; j < (int)v[i].size(); j++)
    {
        if ( v[i][j] == fa ) continue;
        if ( !search(v[i][j], i) ) return false;
        A[i] += A[v[i][j]] * B[i];
        C[i] += C[v[i][j]] * B[i];
        tmp  += B[v[i][j]] * B[i];
    }
    if ( fabs(tmp - 1) < 1e-10 ) return false;
    A[i] /= 1 - tmp;
    B[i] /= 1 - tmp;
    C[i] /= 1 - tmp;
    return true;
}
 
int main()
{
    int nc, n, s, t;
 
    cin >> nc;
    for (int ca = 1; ca <= nc; ca++)
    {
        cin >> n;
        for (int i = 1; i <= n; i++)
            v[i].clear();
 
        for (int i = 1; i < n; i++)
        {
            cin >> s >> t;
            v[s].push_back(t);
            v[t].push_back(s);
        }
        for (int i = 1; i <= n; i++)
        {
            cin >> k[i] >> e[i];
            k[i] /= 100.0;
            e[i] /= 100.0;
        }
        
        cout << "Case " << ca << ": ";
        if ( search(1, -1) && fabs(1 - A[1]) > 1e-10 )
            cout << C[1]/(1 - A[1]) << endl;
        else
            cout << "impossible" << endl;
    }
    return 0;
}